# Invertible matrix / determinant

• Wildcat
In summary, you proved that if B is invertible, then there exists a scalar cεC s.t. A + cB is not invertible.

## Homework Statement

Let A,B ε Mnxn(C)
Prove that if B is invertible, then there exists a scalar cεC s.t. A + cB is not invertible. HINT:
examine det (A+cB)

## The Attempt at a Solution

I know that the det(A+cB)=0 since a non invertible matrix has det=0
I know B is invertible so I multiply the right side by B^-1 which gives det(AB^-1 + cIn)=0 where In is the nxn identity matrix. There is a theorem that says that a scalar is an eigenvalue of A if and only if det(A-cIn)=0.
can I choose c to be -c? does that show that A +cB is not invertible? Help!

So you need to find a c such that $$\det(AB^{-1}+cI_n )=0$$. So this amounts to finding an eigenvalue c' of $$AB^{-1}$$ (or showing that such one exists) and setting c=-c'...

You are on the right track. I think you are using the same letters to represent too many things. Do you care about the eigenvalues of A or of another matrix? Choosing c to be -c would imply that c=0, perhaps you need to use another symbol to represent the quantity that you'd use to choose your original c.

can I rename AB^-1, D and say det(D-cIn)=0 I can find c for a specific 2x2 matrix, but I'm not sure how to do that in general

Wildcat said:
can I rename AB^-1, D and say det(D-cIn)=0 I can find c for a specific 2x2 matrix, but I'm not sure how to do that in general

You're using c for too many things, it's better to write det(D-dIn)=0. The nature of the problem only requires that you know that at least one d exists, not that you have to find its value. Does every matrix have at least one eigenvalue?

Yes? and an nxn would have at most n distinct eigenvalues right?

Wildcat said:
Yes? and an nxn would have at most n distinct eigenvalues right?

Right, so you had the right idea before, you just want to express c in terms of one of the eigenvalues d so that the notation isn't confusing.

so, if I use my original proof and add Let AB^-1 = D and c=-d then det(D-dIn)=0 would be enough? I don't have to show anything else?

Wildcat said:
so, if I use my original proof and add Let AB^-1 = D and c=-d then det(D-dIn)=0 would be enough? I don't have to show anything else?

You can always follow your logic backwards to check that if c = -d, det(A+cB)=0. But your arguments looked fine to me.

OK, Thank You!